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(A) The equation of a plane passing through the point $(x_1, y_1, z_1)$ is given by $A(x - x_1) + B(y - y_1) + C(z - z_1) = 0$. Substituting the point

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$5x - 4y - z = 7$

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(A) The equation of a plane passing through the point $(x_1, y_1, z_1)$ is given by $A(x - x_1) + B(y - y_1) + C(z - z_1) = 0$. Substituting the point $(1, -1, 2)$,we get $A(x - 1) + B(y + 1) + C(z - 2) = 0$ ... $(1)$. Since this plane is perpendicular to the planes $2x + 3y - 2z = 5$ and $x + 2y - 3z = 8$,the normal vector of our plane $(A, B, C)$ must be perpendicular to the normals of the given planes $(2, 3, -2)$ and $(1, 2, -3)$. Thus,the normal vector is the cross product of $(2, 3, -2)$ and $(1, 2, -3)$: $\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & -2 \ 1 & 2 & -3 \end{vmatrix} = \hat{i}(-9 - (-4)) - \hat{j}(-6 - (-2)) + \hat{k}(4 - 3) = -5\hat{i} + 4\hat{j} + 1\hat{k}$. So,$A = -5, B = 4, C = 1$. Substituting these into $(1)$: $-5(x - 1) + 4(y + 1) + 1(z - 2) = 0$. $-5x + 5 + 4y + 4 + z - 2 = 0$. $-5x + 4y + z + 7 = 0$,which simplifies to $5x - 4y - z = 7$.

Find the equation of the plane that contains the point $(1, -1, 2)$ and is perpendicular to each of the planes $2x + 3y - 2z = 5$ and $x + 2y - 3z = 8$.

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